K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
8 tháng 2 2020
Bài 2 :
a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)
c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)
=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)
d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)
e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)
=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)
f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)
=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)
8 tháng 2 2020
Bài 1 :
a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)
=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)
=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)
=> \(12-3x-9-2x+4=0\)
=> \(-5x=-7\)
=> \(x=\frac{7}{5}\)
HH
7 tháng 7 2018
a/ \(\left(5x+1\right)^2=\left(3x-2\right)^2\)
<=> \(\left(5x+1\right)^2-\left(3x-2\right)^2=0\)
<=> \(\left(5x+1-3x+2\right)\left(5x+1+3x-2\right)=0\)
<=> \(\left(2x+3\right)\left(8x-3\right)=0\)
<=> \(\orbr{\begin{cases}2x+3=0\\8x-3=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{3}{2}\\x=\frac{3}{8}\end{cases}}\)
AK
7 tháng 7 2018
a )
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Rightarrow\left(5x\right)^2+2.5x.1+1=\left(3x\right)^2-2.3x.2+2^2\)
\(\Rightarrow25x^2+10x+1=9x^2-12x+4\)
\(\Rightarrow25x^2+10x+1-9x^2+12x-4=0\)
\(\Rightarrow16x^2+22x-3=0\)
\(\Rightarrow\left(4x\right)^2+2.4x.2,75+\left(2,75\right)^2-10,5625=0\)
\(\Rightarrow\left(4x+2,75\right)^2=10,5625\)
\(\Rightarrow4x+2,75=3,25\)
\(\Rightarrow4x=0,5\)
\(\Rightarrow x=0,125\)
Vậy \(x=0,125\)
ND
7 tháng 7 2018
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
ND
7 tháng 7 2018
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
KS
2
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 2 2020
\(\Leftrightarrow x^3+x^2-2x+5x^2+5x-10=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)+5\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+2\right)\left(x-1\right)=0\)
b/ \(\Leftrightarrow x^3+5x^2+6x-x^2-5x-6=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
NH
10 tháng 2 2020
\(x^3+6x^2+3x-10=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+10x-10=0\)
\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+5x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-5\end{matrix}\right.\)
Vậy \(S=\left\{1;-2;-5\right\}\)
\(x^3+4x^2+x-6=0\)
\(\Leftrightarrow x^3-x^2+5x^2-5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{1;-2;-3\right\}\)
31 tháng 1 2020
Câu c : \(x^4-3x^3+2x^2-9x+9=0\)
<=>\(x^4-x^3-2x^3+2x^2-9x+9=0\)
<=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x^3-2x^2-9\right)=0\)
<=> \(x-1=0\) hoặc \(x^3-2x^2-9=0\)
Nếu x-1=0 <=> x=1
Nếu \(x^3-2x^2-9=0\)
<=> \(x^3-3x^2+x^2-9=0\)
<=>\(x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
<=>\(\left(x-3\right)\left(x^2+x+3\right)=0\)
Vì \(x^2+x+3=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\) >0 nên x-3=0 <=> x=3
Vậy \(S=\left\{1;3\right\}\)
31 tháng 1 2020
Câu b : \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)
<=> \(4x^2\left(x^2+2x+2\right)=5\left(x^2+2x+1\right)\)
<=> \(4x^4+8x^3+8x^2=5x^2+10x+5\)
<=>\(4x^4+8x^3+3x^2-10x-5=0\)
<=>\(4x^4-4x^3+12x^3-12x^2+15x^2-15x+5x-5=0\)
<=>\(\left(x-1\right)\left(4x^3+12x^2+15x+5\right)=0\)
<=>\(\left(x-1\right)\left(2x+1\right)\left(2x^2+5x+5\right)=0\)
<=>x=1 hoặc \(x=\frac{-1}{2}\)
Phương trình \(2x^2+5x+5=0\) Vô nghiệm
MV
0
Q
6 tháng 7 2017
câu 5: đặt x2 = t, khi đó:
\(-x^4+2x^2+1=0\) (5)
\(\Leftrightarrow-t^2+2t+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\\t=1-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=1+\sqrt{2}\\x^2=1-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{1+\sqrt{2}}\\x=-\sqrt{1+\sqrt{2}}\\x\in R\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{1+\sqrt{2}}\\x=-\sqrt{1+\sqrt{2}}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (5) là \(S=\left\{-\sqrt{1+\sqrt{2}};\sqrt{1+\sqrt{2}}\right\}\)
\(x^4+x^3+\frac{1}{4}x^2+3x^2-3x+\frac{3}{4}+\frac{3}{4}x^2+\frac{17}{4}=0\)
\(\Leftrightarrow x^2\left(x^2+x+\frac{1}{4}\right)+3\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}x^2+\frac{17}{4}=0\)
\(\Leftrightarrow x^2\left(x+\frac{1}{2}\right)^2+3\left(x-\frac{1}{2}\right)^2+\frac{3}{4}x^2+\frac{17}{4}=0\)
Phương trình vô nghiệm